An organ pipe P_1 closed at one end vibrating | vedclass

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(C) For a closed organ pipe of length $L_1$,the frequency of the $k$-th overtone is given by $n = \frac{(2k+1)v}{4L_1}$. For the first overtone,$k=1$,

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$3:8$

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(C) For a closed organ pipe of length $L_1$,the frequency of the $k$-th overtone is given by $n = \frac{(2k+1)v}{4L_1}$. For the first overtone,$k=1$,so $n_1 = \frac{3v}{4L_1}$.For an open organ pipe of length $L_2$,the frequency of the $m$-th overtone is given by $n = \frac{(m+1)v}{2L_2}$. For the third overtone,$m=3$,so $n_2 = \frac{4v}{2L_2} = \frac{2v}{L_2}$.Since both pipes are in resonance with the same tuning fork,$n_1 = n_2$.Equating the two: $\frac{3v}{4L_1} = \frac{2v}{L_2}$.Rearranging for the ratio $\frac{L_1}{L_2}$: $\frac{L_1}{L_2} = \frac{3v}{4} 	imes \frac{1}{2v} = \frac{3}{8}$.Thus,the ratio is $3:8$.

An organ pipe $P_1$ closed at one end vibrating in its first overtone and another pipe $P_2$ open at both ends vibrating in its third overtone are in resonance with a given tuning fork. The ratio of lengths of $P_1$ and $P_2$ is

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